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I am just getting into this software. I construct my circuit, and it tells me to choose a ground. I only have experience with "ideal" circuits and circuit elements, and have only seen "ground" in the context of the node-voltage method for solving circuits. My question is: What does this term "choose a ground" mean, and where should I put it in a simple circuit with a voltage source, a resistor and an LED all in series?

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  • \$\begingroup\$ Very loosely speaking, "choose a ground" means "choose the point where the black lead of your voltmeter will be connected". As Brian correctly points out, that node is, by definition, the the reference node to which all the node voltages in the simulation are referenced. \$\endgroup\$ – Alfred Centauri Apr 30 '13 at 21:32
  • \$\begingroup\$ Well, loosely speaking, you are in the context of something resembling the node-voltage method for solving circuits!!! You have to teach the software where your idea of 0V is, otherwise some of the numbers that come out of the simulation (absolute voltages) cannot be meaningfully reported. It would be nice if the software made an educated guess for a default choice of ground (like CircuitLab does!) \$\endgroup\$ – Kaz May 1 '13 at 0:09
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Ground is a notional "0V" point : all voltages are referenced to it. In principle it could be applied to any node in such a simple circuit; however the usual convention is to choose the -ve terminal of the voltage source (or any point directly connected to it, i.e. on the same net). Any other choice runs the risk of confusing people!

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